Question

Light of wavelength 350. nm falls on a potassium surface. Electrons are emitted with kinetic energy of 1.31 eV. Find a. The work function of potassium b. The cutoff wavelength c. The frequency corresponding to the cutoff wavelength.

Answer 1

(b)The threshold wavelength is the value of maximum wavelength that is needed to remove an electron from the outermost shell of an isolated atom.

(c)Part(c) :

f=cλ

c=velocity of light=3×108 m/s.

λ=5.6×10−7 m

Plugging the values in the above formula we get:

f=3×1085.6×10−7=5.4×1014 Hz

λT=h×cW=6.6×10−34×3×1082.23×1.6×10−19=5.6×10−7 m .

Step-by-step explanation:

According to the Einstein law of photoelectric effect, when any radiation strikes a surface some of its energy is used in taking an electron out of that atom and remaining energy is converted in the form of kinetic energy of that electron.